Mixtures and compounds

Mixtures and compounds represent one of the more tricky set of questions in the word problem section of the IPAT.  As always, let’s use a few progressively harder examples to clear the underlying concepts tested in the IPAT.

Q) A solution x contains 30% chlorine and another solution y contains 50% chlorine. How many liters of each need to be mixed to generate a solution containing 40% chlorine and comprising a total of 10 liter in volume.

Whenever you come across a mixtures problem, start thinking by how you can represent the given data by a series of linear equations

Since, we need the final breakdown of each solution to form the compound solution, lets start representing them as X liters of x and Y liters of y.

Clearly, we want the sum of the two solution to be exactly 10 liters

X + Y = 10

Now, the chlorine equivalence equation, 30% of X + 50% of Y = 40% of final solution (10 liters)

0.3X + 0.5Y = 0.4(10) = 4

Solving the pair of equations we get, X = 5 liters and Y = 5 liters

Q) Two solutions A and B, cost in the proportion 2:3. If a mixture containing 20% of solution A and rest of B, sells at a 20% profit margin at the price of $300. What is the cost of solution B?

Solution

Let’s start by finding the total cost of the solution

(Price – Cost)x100/Price = profit margin (in percentage)

(300 – Cost)x100/300 = 20

300 – Cost = 60

Cost = $240

Now, if the cost of solution A is 2x/litre, the cost of solution B is 3x/litre (as they are in the ratio 2:3)

Let say the total volume of the solution is V, therefore volume of solution A is 0.2V and volume of solution B is 0.8V (as 20% of mixture by volume is A)

Therefore cost of solution A = (price of A/litre)xvolume of A = 2x(0.2V) = 0.4xv

Similarly cost of solution B = 3x(0.8V) = 2.4xv

Total cost = 0.4xv + 2.4xv = 240

Therefore xv = $85.7

Cost of solution B in compound solution is 2.4xv = (2.4)85.7 = $205.7 ~ $206 (appx)

 

Q) Solution X contains 30% of solution B, 20% solution T and the rest is water. 3 kg of water and 1 kg of solution T evaporates from 10 Kg of the combined solution. If 2 kg of the orginial mixture combination is added to the solution remaining after the evaporation process,what percentage of the solution is B?

Solution

Process type questions are common in the IPAT. You have an original solution, a process(1) happens, another process(2) happens in some time, now what are the remaining constituents? It is recommended you work sequentially for these type of problems.

Let’s start by dividing the original solution into its constituent parts by weight.

Weight of Solution B = 0.3(10) = 3 Kg

Weight of Solution T = 0.2(10) = 2 Kg

Weight of Water = 10 – 3 -2 = 5 Kg

Process 1 Evaporation Process

We lose, 3 Kg of water and 1 Kg of solution T

Therefore, new weight of solution T = 2 – 1 = 1Kg

New weight of water = 5Kg – 3Kg = 2Kg

New mixture constituents

Weight of Solution B = 0.3(10) = 3 Kg

Weight of Solution T = 0.2(10) = 1 Kg

Weight of Water = 10 – 3 -2 = 2 Kg

Process 2 Addition of original solution

2 Kg of original solution is now added, this maps to (0.3)2 = 0.6 Kg of solution B, 0.2(2) = 0.4Kg of solution T and the rest of water (2-0.6-0.4 = 1 Kg)

Updated mixture constituents

Weight of Solution B = 0.3(10) = 3 + 0.6 = 3.6 Kg

Weight of Solution T = 0.2(10) = 1 + 0.4 = 1.4 Kg

Weight of Water = 10 – 3 -2 = 2 + 1= 3 Kg

Therefore percentage of solution B = 3.6/(3.6+1.4+3) = 45%

 

If you found this useful and want a more extensive coverage of IPAT mixtures and compounds questions check out our comprehensive practice guide on IPAT word problems here.

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